In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to :
- Prevent interference by phosphate ions
- Decrease concentration of \( \text{OH}^- \) ions
- Increase concentration of \( \text{Cl}^- \) ions
- Increase concentration of \( \text{NH}_4^+ \) ions
The Correct Option is B
Approach Solution - 1
Role of Ammonium Chloride in Precipitation of Group III Cations:
In qualitative analysis, group III cations (such as \( \text{Fe}^{3+}, \text{Cr}^{3+}, \text{Al}^{3+} \)) are precipitated as hydroxides by adding ammonium hydroxide \( \text{NH}_4\text{OH} \).
Effect of Adding Ammonium Chloride:
Ammonium chloride \( \text{NH}_4\text{Cl} \) is added before ammonium hydroxide to control the concentration of \( \text{OH}^- \) ions. This is achieved through the common ion effect:
\[ \text{NH}_4\text{OH} \leftrightarrow \text{NH}_4^+ + \text{OH}^- \]
Adding \( \text{NH}_4\text{Cl} \) increases the concentration of \( \text{NH}_4^+ \) ions, which shifts the equilibrium to the left, decreasing the concentration of \( \text{OH}^- \) ions.
Importance of Controlled \( \text{OH}^- \) Concentration:
By reducing \( \text{OH}^- \) concentration, we avoid the formation of precipitates from cations of higher groups (such as Group IV and V cations), ensuring selective precipitation of Group III cations only.
Conclusion:
The addition of ammonium chloride decreases the concentration of \( \text{OH}^- \) ions through the common ion effect, which corresponds to Option (2).
Approach Solution -2
In qualitative inorganic analysis, the iron group (Group III) cations such as \( \text{Fe}^{3+} \), \( \text{Al}^{3+} \), and \( \text{Cr}^{3+} \) are separated by precipitating as their hydroxides.
Step 2: Role of Ammonium Hydroxide (\( \text{NH}_4\text{OH} \)) in Precipitation.
Adding ammonium hydroxide provides \( \text{OH}^- \) ions which react with metal ions to form insoluble metal hydroxides, thus precipitating the iron group cations.
Step 3: Why Ammonium Chloride (\( \text{NH}_4\text{Cl} \)) is Added Before Ammonium Hydroxide.
- Ammonium chloride dissociates in solution to give ammonium ions (\( \text{NH}_4^+ \)) and chloride ions (\( \text{Cl}^- \)).
- The presence of \( \text{NH}_4^+ \) ions shifts the equilibrium:
\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \] towards the left due to the common ion effect.
- This results in a decreased concentration of free \( \text{OH}^- \) ions in solution.
Step 4: Purpose of Controlling \( \text{OH}^- \) Concentration.
- If \( \text{OH}^- \) concentration is not controlled and is too high, cations from other groups (like Group IV and V) may also precipitate.
- Reducing \( \text{OH}^- \) concentration ensures selective precipitation of iron group hydroxides only, avoiding contamination with other metal hydroxides.
Step 5: Summary and Conclusion.
Therefore, ammonium chloride is added before ammonium hydroxide in the precipitation of the iron group (III) in qualitative analysis to decrease the concentration of hydroxide ions (\( \text{OH}^- \)) in the solution. This helps in selective precipitation of only the iron group cations.
Final Answer:
\[ \boxed{ \text{Decrease concentration of } \text{OH}^- \text{ ions} } \]
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Qualitative Analysis Questions
- Given below are two statements:
Statement-I: The gas liberated on warming a salt with dil H2SO4, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.
Statement-II: In statement-I the colour of paper turns black because of formation of lead sulphite. In the light of the above statements, choose the most appropriate answer from the options given below: - The Lassaigne’s extract is boiled with dil HNO\(_3\) before testing for halogens because
- In chromyl chloride test for confirmation of Cl-, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2 turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is
- A reagent which gives brilliant red precipitate with Nickel ions in basic medium is
- On passing a gas, ‘X’, through Nessler’s reagent, a brown precipitate is obtained. The gas ‘X’ is
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.